If value of $\int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx} = \frac{A}{sin\alpha} + B (\alpha \ne 0) $ the values of A and B are

Q: If value of $\displaystyle \int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx} = \frac{A}{sin\alpha} + B (\alpha \ne 0) $ the values of A and B are

(a) $\displaystyle A =\frac{\pi}{2} , B = 0 $

(b) $\displaystyle A =\frac{\pi}{4} , B = \frac{\pi}{4 sin\alpha} $

(c) $\displaystyle A =\frac{\pi}{6} , B = \frac{\pi}{ sin\alpha} $

(d) $\displaystyle A =\pi , B = \frac{\pi}{ sin\alpha} $

Ans: (a) , (b)

Sol: $\displaystyle I = \int_{0}^{\alpha} \frac{dx}{1-cos\alpha cosx} $

$\displaystyle I = \int_{0}^{\alpha} \frac{dx}{cos^2 \frac{x}{2} + sin^2 \frac{x}{2}-cos\alpha (cos^2 \frac{x}{2} – sin^2 \frac{x}{2})} $

$\displaystyle I = \int_{0}^{\alpha} \frac{dx}{(1-cos\alpha) cos^2 \frac{x}{2} + (1 + cos\alpha) sin^2 \frac{x}{2}} $

$\displaystyle I = \int_{0}^{\alpha} \frac{sec^2 \frac{x}{2} dx}{2 sin^2 \frac{\alpha}{2} + 2 cos^2 \frac{\alpha}{2} tan^2 \frac{x}{2}} $

$\displaystyle I = \frac{1}{2} \int_{0}^{\alpha} \frac{sec^2 \frac{\alpha}{2} sec^2 \frac{x}{2} dx}{tan^2 \frac{\alpha}{2} + tan^2 \frac{x}{2}} $

Let $tan\frac{x}{2} = t $

$\displaystyle I = \int_{0}^{tan(\alpha/2)} \frac{sec^2 \frac{\alpha}{2} dt }{t^2 + tan^2 \frac{\alpha}{2}} $

$\displaystyle I = sec^2 \frac{\alpha}{2} cot\frac{\alpha}{2} [tan^{-1}\frac{t}{tan(\alpha/2)}]_{0}^{tan(\alpha/2)} $

$\displaystyle = \frac{2}{sin\alpha}.\frac{\pi}{4}= \frac{\pi}{2 sin\alpha}$

$\displaystyle A =\frac{\pi}{2} , B = 0 $

Also , $\displaystyle A =\frac{\pi}{4} , B = \frac{\pi}{4 sin\alpha} $

Hence (a), (b) are correct answers.

The value of $\int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x} $ is

Q: The value of $\displaystyle \int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x} $ is

(A) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{b}{a}) ; a > 0 , b > 0$

(B) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{b}{a}) ; a < 0 , b < 0$

(C) $\displaystyle \frac{\pi}{4} ; a = 1 , b = 1 $

(D) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{a}{b}) + \frac{1}{ab}$

Ans: (A) , (B) , (C)

Sol: $\displaystyle I = \int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x} $

Dividing Nr and Dr by cos2x ,

$\displaystyle I = \int_{0}^{\pi/4} \frac{sec^2 x dx}{a^2 + b^2 tan^2 x} $

Put tanx = t ; sec2x dx = dt

$\displaystyle I = \int_{0}^{1} \frac{dt}{a^2 + b^2 t^2} $

$\displaystyle I = \frac{1}{b^2} \int_{0}^{1} \frac{dt}{t^2 + \frac{a^2}{b^2}} $

$\displaystyle I = \frac{1}{b^2} \int_{0}^{1} \frac{dt}{t^2 + (\frac{a}{b})^2} $

$\displaystyle = \frac{b}{a b^2} [tan^{-1}\frac{b t}{a}]_{0}^{1} $

$\displaystyle = \frac{1}{a b} [tan^{-1}\frac{b}{a}] $ ; (a ≠0 , b≠ 0)

Also for a = 1, b = 1

$\displaystyle I = \int_{0}^{\pi/4}dx = \frac{\pi}{4} $

Hence (A), (B), (C) are correct answers.

If $ f(x) = \int_{x}^{x^2} \frac{dt}{(logt)^2} , x \ne 0$ then f(x) is

Q: If $\displaystyle f(x) = \int_{x}^{x^2} \frac{dt}{(logt)^2} , x \ne 0$ then f(x) is

(A) monotonically increasing in (2 , ∞)

(B) monotonically decreasing in (1 , 2)

(C) monotonically decreasing in (2 , ∞)

(D) monotonically decreasing in (0 , 1)

Ans: (A) , (B)

Sol: $\displaystyle f'(x) = \frac{2x}{(logx^2)^2} – \frac{1}{(logx)^2}$

$\displaystyle f'(x) = \frac{x-2}{2(logx)^2} $

f'(x) > 0 for x > 2

and f'(x) < 0 for x < 2

Hence (A), (B) are correct answers.

Find all the possible values of the parameter ‘a ‘ so that the function …

Q: Find all the possible values of the parameter ‘a ‘ so that the function,
f (x) = x3 – 3 (7 – a) x2 – 3 (9- a2) x + 2 , has a negative point of local minimum.

Sol. f(x) = x3 –3(7-a) x2 –3(9-a2)x + 2

f ‘(x) = 3x2 –6(7-a) x – 3(9-a2)

For distinct real roots D > 0

36(7-a)2 + 4 ×3×3 (9 – a2) > 0

⇒ 49 + a2 –14a + 9 –a2 > 0

14a < 58 ⇒ a < 29/7 For local minima f ”(x) = 6x – 6(7 – a) > 0

⇒ x – 7 + a > 0

7 – a < x as x must be -ve

⇒ 7-a < 0 ⇒ a > 7

Thus, by contradiction i.e. for real roots a < and for negative point of local minimum a > 7. No possible value of a.

$ Let \; f(x) = \left\{\begin{array}{ll} -x^3 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \; , 0 \leq x < 1 \\ 2x - 3 \; , 1 \leq x \leq 3 \end{array} \right. $

Q: $ \large Let \; f(x) = \left\{\begin{array}{ll} -x^3 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \; , 0 \leq x < 1 \\ 2x – 3 \; , 1 \leq x \leq 3 \end{array} \right. $

Find all possible real values of b such that f(x) has the smallest value at x = 1.

Sol. At x = 1, f(x) = -1

⇒ Smallest value of f(x) = -1

⇒ at all other points of the interval, f(x) > -1.

Now, for x ≥ 1, f(x) = 2x –3

⇒ f ‘(x) = 2 > 0 ⇒ f(x) is an increasing function

⇒ least value exists at x =1.

Again, for x < 1, f ‘(x) = -3x2 < 0

⇒ f(x) is decreasing function in the interval 0 ≤ x < 1.

Therefore, f(x) is smallest at x =1 provided

$\displaystyle f(1-0) = \lim_{h\rightarrow 0} -(1-h)^3 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \geq -1$

$\displaystyle -1 + \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \geq -1$

$\displaystyle \frac{b^3 – b^2 + b – 1}{b^2 + 3b +2} \geq 0 $

$\displaystyle \frac{(b^2 + 1)(b – 1}{(b + 1)(b +2 )} \geq 0 $

$\displaystyle \frac{(b – 1}{(b + 1)(b +2 )} \geq 0 $ ; Since (b^2 + 1) is always +ve

⇒ b ∈ (-2 ,-1) ∪ [ 1 , ∞)